C++ 中列表中的缺失排列
问题陈述
给定任何单词的排列列表。从排列列表中找出缺失的排列。
示例
If permutation is = { “ABC”, “ACB”, “BAC”, “BCA”} then missing permutations are {“CBA” and “CAB”}
算法
- 创建所有给定字符串的集合
- 以及所有排列的另一个集合
- 返回两个集合之间的差异
示例
#include <bits/stdc++.h> using namespace std; void findMissingPermutation(string givenPermutation[], size_t permutationSize) { vector<string> permutations; string input = givenPermutation[0]; permutations.push_back(input); while (true) { string p = permutations.back(); next_permutation(p.begin(), p.end()); if (p == permutations.front()) break; permutations.push_back(p); } vector<string> missing; set<string> givenPermutations(givenPermutation, givenPermutation + permutationSize); set_difference(permutations.begin(), permutations.end(), givenPermutations.begin(), givenPermutations.end(), back_inserter(missing)); cout << "Missing permutations are" << endl; for (auto i = missing.begin(); i != missing.end(); ++i) cout << *i << endl; } int main() { string givenPermutation[] = {"ABC", "ACB", "BAC", "BCA"}; size_t permutationSize = sizeof(givenPermutation) / sizeof(*givenPermutation); findMissingPermutation(givenPermutation, permutationSize); return 0; }
在编译和执行以上程序时。它生成以下输出 −
输出
Missing permutations are CAB CBA
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